Lemma: Let $A$ be a $2 \times 2$ matrix over a field $F$. If $A$ is not a scalar multiple of the identity matrix, then $A$ is similar to
\begin{pmatrix}
0 & -d\\
1 & t
\end{pmatrix}
For a linear transformation $T: (\mathbb{Z}_p \times \mathbb{Z}_p) \rightarrow (\mathbb{Z}_p \times \mathbb{Z}_p)$, we know that $T \in GL_2(\mathbb{Z}_p)$. $T^q=I$, where $q<p$ and $p,q >2$, $p$ and $q$ are distinct primes. This implies that the minimal polynomial $m(x)$ for $T$ divides $x^q-1$.
Moreover, there exist some non-zero vector $v \in (\mathbb{Z}_p \times \mathbb{Z}_p)$ which is not an eigen vector of $T$ and the pair $\{v, T(v)\}$ forms a basis of $(\mathbb{Z}_p \times \mathbb{Z}_p)$. The matrix of $T$ with respect to this basis has the form
\begin{pmatrix}
0 & a\\
1 & b
\end{pmatrix}.
This is similar to $A$ (i.e. a=-d, b=t).Based on the above information regarding $T$ and $p,q$, can I determine possible values for $a,b$? In other words, how can I determine possible values for $a,b$ such that $T$ represents a generator of an order $q$ subgroup in $GL_2(\mathbb{Z}_p)$?
$q$ can divide $p^2-1$. i.e. $q|(p-1)$ or $q|(p+1)$.
Best Answer
Let $\Bbb F_{p^2}$ be the quadratic extension of $\Bbb F_p$ and let $\langle\lambda\rangle=\Bbb F_{p^2}^*$. Presumably, $q\;\Big\vert\;|\Bbb F_{p^2}^*|=p^2-1$ and therefore $\Bbb F_{p^2}^*$ contains a size $q$ cyclic subgroup. Let $a=-\alpha\bar{\alpha}\;,\;b=\alpha+\bar{\alpha}$ so that
$$\mathbf A=\begin{pmatrix}0&-\alpha\bar{\alpha}\\1&\alpha+\bar{\alpha}\\\end{pmatrix}$$ where $\alpha=\lambda^{\frac{p^2-1}{q}}$. Now, $\mathbf A,\mathbf D$ are similar as they have the same characteristic polynomial $$\chi(x)=x^2-(\alpha+\bar{\alpha})x+\alpha\bar{\alpha}$$ Therefore, $\langle\mathbf A\rangle,\langle\mathbf D\rangle$ are conjugate size $q$ cyclic $\text{GL}(2,p^2)$ subgroups but with $\langle\mathbf A\rangle\subseteq\text{GL}(2,p)$.